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-ln x = t/50 + C Solve for x
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\[-\ln x = \left(\begin{matrix}t \\ 50\end{matrix}\right)+ C\]
\[ x = -e^{t/50} - e^c\]
Is the e when you divie the - ln ?
no, to get x out of ln, we take the e of ln x, so e and ln canceles, leaving x on left side on the right side, every term becomes exponent of the e
Do I add the exponents and leave a postive e? Answer: \[e ^{-C-t/50}\]
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yes you can simplify like that
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